Practical flow quantities [Fluid Mechanics

[@darksca1804]

Professor, at 10:03, why is the pressure at point A equal to rho*g*h (the gauge pressure) only? Shouldn't it be the sum of the ambient pressure and rho*g*h (the absolute pressure at point A)?

[@prof.vanburen]

Good question and sorry to be so late with a reply! Technically the pressure at that point has an ambient pressure contribution and it isn't exactly correct to not write it down. However, the pressure at the top and bottom of the device are equal, so it does not end up contributing to the driving pressure of the system and can be ignored.

[@bluemantle8269]

You are skipping a lot of steps and coz of tat its hard to understand, especially at 7:51 you just say "u" reduces to following equation and now I'm confused hw tat happened. How did dh upon dx became negative 1

[@prof.vanburen]

At that specific time the only operation is distributing the d/dx into the parenthesis and recognizing that dh/dx = -1. This is because h and x are on the same axis and opposite directions, so a change in x is a negative equivalent change in h.